Appendix: Mathematical Details

### Schwarzschild Curvature

The Schwarzschild line element is $$ds^2 = -f\,dt^2 + \frac{dr^2}{f} + r^2\,d\theta^2 + r^2\sin^2\theta\,d\phi^2$$ where we have written $f$ for the function $$f(r) = 1-\frac{2m}{r}$$ with derivatives \begin{align} f' &= \frac{df}{dr} = \frac{2m}{r^2} \\ \fpp &= -\frac{4m}{r^3} \end{align} It now follows immediately that $$d\rr = \sqrt{f}\,dt\,\That + \frac{dr}{\sqrt{f}}\,\rhat + r\,d\theta\,\that + r\,\sin\theta\,d\phi\,\phat$$ and the basis 1-forms are \begin{align} \sigma^t &= \sqrt{f}\,dt \\ \sigma^r &= \frac{dr}{\sqrt{f}} \\ \sigma^\theta &= r\,d\theta \\ \sigma^\phi &= r\,\sin\theta\,d\phi \end{align} The structure equations therefore become \begin{align} d\sigma^t &= \frac12 \frac{f'}{\sqrt{f}}\>dr\wedge dt = -\omega^t{}_m\wedge\sigma^m \\ d\sigma^r &= 0 = -\omega^r{}_m\wedge\sigma^m \\ d\sigma^\theta &= dr\wedge d\theta = -\omega^\theta{}_m\wedge\sigma^m \\ d\sigma^\phi &= \sin\theta\,dr\wedge d\phi + r\,\cos\theta\,d\theta\wedge d\phi = -\omega^\phi{}_m\wedge\sigma^m \end{align} These equations suggest that \begin{align} \omega^t{}_r &= \frac12 f'\,dt = \frac12 \frac{f'}{\sqrt{f}}\,\sigma^t \\ \omega^\theta{}_r &= \sqrt{f}\,d\theta = \frac{\sqrt{f}}{r}\,\sigma^\theta \\ \omega^\phi{}_r &= \sqrt{f}\,\sin\theta\,d\phi = \frac{\sqrt{f}}{r}\,\sigma^\phi \\ \omega^\phi{}_\theta &= \cos\theta\,d\phi = \frac{1}{r}\cot\theta\,\sigma^\phi \end{align} and it is easy to check that these educated guesses actually do satisfy the structure equations, remembering that \begin{align} \omega^r{}_t &= \omega^t{}_r, & \omega^r{}_\theta &= -\omega^\theta{}_r, & \omega^r{}_\phi &= -\omega^\phi{}_r, & \omega^\theta{}_\phi = -\omega^\phi{}_\theta \end{align} but setting all other connection 1-forms to zero. Since we are guaranteed a unique solution, we have found our connection 1-forms.

The curvature 2-forms are now given by \begin{align} \Omega^t{}_r &= d\omega^t{}_r + \omega^t{}_m\wedge\omega^m{}_r = \frac12 \fpp\,dr\wedge dt = -\frac12 \fpp\,\sigma^t\wedge\sigma^r \\ \Omega^t{}_\theta &= d\omega^t{}_\theta + \omega^t{}_m\wedge\omega^m{}_\theta = - \frac{f'}{2}\sqrt{f}\,dt\wedge d\theta = -\frac{f'}{2r} \sigma^t\wedge\sigma^\theta \\ \Omega^t{}_\phi &= d\omega^t{}_\phi + \omega^t{}_m\wedge\omega^m{}_\phi \nonumber\\ & = - \frac{f'}{2}\sqrt{f}\,\sin\theta\,dt\wedge d\phi = -\frac{f'}{2r} \sigma^t\wedge\sigma^\phi \\ \Omega^\theta{}_r &= d\omega^\theta{}_r + \omega^\theta{}_m\wedge\omega^m{}_r = \frac12 \frac{f'}{\sqrt{f}}\,dr\wedge d\theta = -\frac{f'}{2r}\,\sigma^\theta\wedge\sigma^r \\ \Omega^\phi{}_r &= d\omega^\phi{}_r + \omega^\phi{}_m\wedge\omega^m{}_r \nonumber\\ &= \frac12 \frac{f'}{\sqrt{f}}\,\sin\theta\,dr\wedge d\phi +\sqrt{f}\,\cos\theta\,d\theta\wedge d\phi \nonumber\\ &\qquad\qquad\qquad\qquad +\sqrt{f}\,\cos\theta\,d\phi\wedge d\theta \nonumber\\ &= -\frac{f'}{2r}\,\sigma^\phi\wedge\sigma^r \\ \Omega^\phi{}_\theta &= d\omega^\phi{}_\theta + \omega^\phi{}_m\wedge\omega^m{}_\theta \nonumber\\ &= -\sin\theta\,d\theta\wedge d\phi - f\sin\theta\,d\phi\wedge d\theta = \frac{1-f}{r^2}\sigma^\phi\wedge\sigma^\theta \end{align}

Again we have \begin{align} \Omega^r{}_t &= \Omega^t{}_r, & \Omega^r{}_\theta &= -\Omega^\theta{}_r, & \Omega^r{}_\phi &= -\Omega^\phi{}_r, & \Omega^\theta{}_\phi &= -\Omega^\phi{}_\theta \end{align} as well as $$\Omega^\theta{}_t = \Omega^t{}_\theta, \quad \Omega^\phi{}_t = \Omega^t{}_\phi$$ Substituting for $f$, we obtain the Schwarzschild curvature as given in §Schwarzschild Connection.

The (nonzero, independent) components of the Riemann tensor are therefore given by \begin{align} R^t{}_{rtr} &= -\frac{\fpp}{2} \\ R^t{}_{\theta t\theta} = R^t{}_{\phi t\phi} &= -\frac{f'}{2r} = R^\theta{}_{r\theta r} = R^\phi{}_{r\phi r} \\ R^\phi{}_{\theta\phi\theta} &= \frac{1-f}{r^2} \end{align} from which it follows that the components of the Ricci tensor are 1) \begin{align} R_{tt} &= R^r{}_{trt} + R^\theta{}_{t\theta t} + R^\phi{}_{t\phi t} = \frac{\fpp}{2} + \frac{f'}{2r} + \frac{f'}{2r} = 0 \\ R_{rr} &= R^t{}_{rtr} + R^\theta{}_{r\theta r} + R^\phi{}_{r\phi r} = -\frac{\fpp}{2} - \frac{f'}{2r} - \frac{f'}{2r} = 0 \\ R_{\theta\theta} &= R^t{}_{\theta t\theta} + R^r{}_{\theta r\theta} + R^\phi{}_{\theta\phi\theta} = -\frac{f'}{2r} - \frac{f'}{2r} + \frac{1-f}{r^2} = 0 \\ R_{\phi\phi} &= R^t{}_{\phi t\phi} + R^r{}_{\phi r\phi} + R^\theta{}_{\phi\theta\phi} = -\frac{f'}{2r} - \frac{f'}{2r} + \frac{1-f}{r^2} = 0 \end{align} Thus, the Schwarzschild geometry is indeed a solution of the Einstein vacuum equation, as claimed.

1) The off-diagonal components of the Ricci tensor, such as $R^t{}_r$ are easily seen to be identically zero.